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Unformatted text preview: PHY 317K Homework 6 Solutions 1. B For a circle (like a coin), A = r 2 = d 2 / 4, where d is the diameter. If d changes by d , the area becomes A + A = ( d + d ) 2 / 4 = d 2 / 4 + 2 d d/ 4 + ( d ) 2 / 4. Therefore, A = 2 d d/ 4, if we ignore the third term since d is very small compared to d . Then if we multiply the left side by d/d , we see A = 2( d 2 / 4) d/d , from which it follows: A/A = 2 d/d . Hence, a 0.17% change in d results in a 0.34% change in the area of the coin. 2. C For coefficient of linear expansion , V V = 3 T . Thus, here V = 3 V T = 3(1 . 1 10 5 /C )(4 . 00cm) 3 (57C ) = 0 . 12cm 3 . 3. D The heat capacity at constant volume and the heat capacity at constant pressure have different values because the system does work at constant pressure, but not at constant volume. I.e., when the volume changes, work is done. 4. A The energy needed to melt A is ML , where M is the mass of A and L is latent heat of fusion. The total heat energy that A receives is C A ( T final T A ) + ML , and this must equal the heat energy lost by B : C B ( T B T final ) = C A ( T final T A ) + ML. Solving for T final we find: T final = C A T A + C B T B ML C A + C B 5. A During the time that latent heat is involved in a change of phase the temperature does not change. 6. E Several steps are involved: 1) The ice is raised from 20 C to 0 C: E 1 = 10g(0 . 5cal / g C )(20C ) = 100cal. 2) The ice is melted: E 2 = 10g(80cal / g) = 800cal. 3) The water is heated to 100 C: E 3 = 10g(1 . 00cal / g C )(100C ) = 1000cal. 4) The water is vaporized: E 4 = 10g(540cal / g) = 5400cal. 5) The steam is heated to 130 C: E 5 = 10g(0 . 5cal / g C )(60C ) = 300cal. The sum of the energies required for all steps is 7600 cal. 1 7. C The humidifier heats 15g of water from 20 C to 100 C and then evaporates those 15g each minute. The energy needed for that is: E = mc T + mL V = (15g) (4190J / kg K) (80K) + (15g) (2256kJ / kg) = 39000J where L V is heat of vaporization. 8. B According to the first law of thermodynamics, applied to gas, the increase in the internal energy during the process is equal to the heat input plus the work done on the gas. 9. D The work done by the gas is equal to W = pdV . Because of the direction of the processes, we see that I and II result in negative work, while III, IV, and V result in positive work. In magnitude, the positive work in IV is largest. This is simply basedpositive work....
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This note was uploaded on 01/11/2010 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.
 Spring '07
 KOPP

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